Problem: On the way to school, Karen became fixated on the puzzle game on her phone!

<image>

The game is played as follows. In each level, you have a grid with n rows and m columns. Each cell originally contains the number 0.

One move consists of choosing one row or column, and adding 1 to all of the cells in that row or column.

To win the level, after all the moves, the number in the cell at the i-th row and j-th column should be equal to gi, j.

Karen is stuck on one level, and wants to know a way to beat this level using the minimum number of moves. Please, help her with this task!

Input

The first line of input contains two integers, n and m (1 ≤ n, m ≤ 100), the number of rows and the number of columns in the grid, respectively.

The next n lines each contain m integers. In particular, the j-th integer in the i-th of these rows contains gi, j (0 ≤ gi, j ≤ 500).

Output

If there is an error and it is actually not possible to beat the level, output a single integer -1.

Otherwise, on the first line, output a single integer k, the minimum number of moves necessary to beat the level.

The next k lines should each contain one of the following, describing the moves in the order they must be done:

  * row x, (1 ≤ x ≤ n) describing a move of the form "choose the x-th row". 
  * col x, (1 ≤ x ≤ m) describing a move of the form "choose the x-th column". 



If there are multiple optimal solutions, output any one of them.

Examples

Input

3 5
2 2 2 3 2
0 0 0 1 0
1 1 1 2 1


Output

4
row 1
row 1
col 4
row 3


Input

3 3
0 0 0
0 1 0
0 0 0


Output

-1


Input

3 3
1 1 1
1 1 1
1 1 1


Output

3
row 1
row 2
row 3

Note

In the first test case, Karen has a grid with 3 rows and 5 columns. She can perform the following 4 moves to beat the level:

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In the second test case, Karen has a grid with 3 rows and 3 columns. It is clear that it is impossible to beat the level; performing any move will create three 1s on the grid, but it is required to only have one 1 in the center.

In the third test case, Karen has a grid with 3 rows and 3 columns. She can perform the following 3 moves to beat the level:

<image>

Note that this is not the only solution; another solution, among others, is col 1, col 2, col 3.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int col[105];
int num[105][105];
int k, mm, mn;
int n, m;
int sum;
int row[105];
void check() {
  for (int i = 0; i < m; i++) {
    if (num[0][i] != mm) {
      col[i] = num[0][i] - mm;
      sum += col[i];
    }
  }
  for (int i = 0; i < n; i++) {
    if (num[i][0] != mn) {
      row[i] = num[i][0] - mn;
      sum += row[i];
    }
  }
}
int flag;
void check2() {
  int i, j, tmp;
  tmp = num[0][0] - row[0] - col[0];
  sum += tmp * min(n, m);
  for (i = 0; i < n; i++) {
    for (j = 1; j < m; j++) {
      if (num[i][j] - col[j] - row[i] != tmp) {
        flag = 0;
        return;
      }
    }
  }
  if (n >= m) {
    for (int i = 0; i < m; i++) col[i] += tmp;
  } else
    for (int i = 0; i < n; i++) row[i] += tmp;
}
int main() {
  scanf("%d%d", &n, &m);
  memset(num, 0, sizeof(num));
  memset(col, 0, sizeof(col));
  memset(row, 0, sizeof(row));
  sum = 0;
  flag = 1;
  mm = 505;
  mn = 505;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < m; j++) {
      scanf("%d", &num[i][j]);
      if (i == 0 && num[i][j] < mm) {
        mm = num[i][j];
      }
      if (j == 0 && num[i][j] < mn) mn = num[i][j];
    }
  }
  check();
  check2();
  if (flag == 0)
    printf("-1\n");
  else {
    printf("%d\n", sum);
    for (int i = 0; i < m; i++) {
      if (col[i] != 0) {
        while (col[i]--) {
          printf("col %d\n", i + 1);
        }
      }
    }
    for (int i = 0; i < n; i++) {
      if (row[i] != 0) {
        while (row[i]--) {
          printf("row %d\n", i + 1);
        }
      }
    }
  }
}